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The links pretty spells out the solution already. If you really can't solve it, why don't you post your attempt here and I'll see where you are stuck on.

@Ujjwal Mani Tripathi
–
Split the integral into $\int_0^{x^3} - \int_0^{t^2}$. For the first integral, just apply the fundamental theorem of calculus. For the second theorem, you will get some constant number, so you can ignore it after you differentiate. Is this enough information for you?

@Ujjwal Mani Tripathi
–
It's not necessary, but if you make the change $t = x \cdot \text{ Sh u}$ with $x$ constant, and use $1 + Sh^2(u) = Ch^2(u)$ and $dt = x \cdot \text{Ch u} \space du$, the problemm looks like easy. Tomorrow, I promise you'll have my answer, now I'm very busy, Im' leaving my house. Tomorrow, I promise....

@Ujjwal Mani Tripathi
–
I think the rest proof 2 should be simple, I'll finish it. I promise.... but you have to say me wheteher I'm right or not... It depends also if $x >0$, $x = 0$ or $x < 0$ to concrete the ending. This question is giving me questions and suggestions... For example, does $\int_0^{t} \space dt$ make some sense?If the answer is yes, Is it $t$?why?... I have another proof making the change $u = \sqrt{x^2 + t^2} + t$... Even we can think first this question without the derivative $\frac{d}{dx}$, I mean given $x$ possible concrete constants, $x =1$, $x =2$, $x = 100$... and later, seeing the results, infering the last answer... And I'm also considering Pi's advice: Use fundamental theorem of Calculus... To be continued... Can you finish my 2nd proof, and I tell you what I think about your result? Come on, dare...

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## Comments

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TopNewestHint: Fundamental Theorem of Calculus.Log in to reply

yeah but i am still not able to get the required answer will you please post the solution to this question...

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The links pretty spells out the solution already. If you really can't solve it, why don't you post your attempt here and I'll see where you are stuck on.

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$\dfrac{3x^2}{\sqrt{x^2 + x^12 }}$ it is x^12 .. in the denominator

i am getting the answerLog in to reply

$\int_0^{x^3} - \int_0^{t^2}$. For the first integral, just apply the fundamental theorem of calculus. For the second theorem, you will get some constant number, so you can ignore it after you differentiate. Is this enough information for you?

Split the integral intoLog in to reply

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$t = x \cdot \text{ Sh u}$ with $x$ constant, and use $1 + Sh^2(u) = Ch^2(u)$ and $dt = x \cdot \text{Ch u} \space du$, the problemm looks like easy. Tomorrow, I promise you'll have my answer, now I'm very busy, Im' leaving my house. Tomorrow, I promise....

It's not necessary, but if you make the changeLog in to reply

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Proof 1 (attempt 1).-$\dfrac d{dx} \int_{t^2}^{x^3} \dfrac 1{\sqrt{x^2 + t^2}} \, dt = \dfrac d{dx} \left( \int_{0}^{x^3} \dfrac 1{\sqrt{x^2 + t^2}} \, dt - \int_{0}^{t^2} \dfrac 1{\sqrt{x^2 + t^2}} \, dt \right) =$ Let's call $g(x) = x^3$ and $f(x) = \int_{0}^{x} \dfrac 1{\sqrt{x^2 + t^2}} \, dt$ Then, $\dfrac d{dx} \int_{0}^{x^3} \dfrac 1{\sqrt{x^2 + t^2}} \, dt = \dfrac d{dx} f(g(x)) = \dfrac d{dx} f(x^3) = 3x^2 \cdot ...$ Sorry, I'm getting a knot for myself...

Proof 2.-$\dfrac d{dx} \int_{t^2}^{x^3} \dfrac 1{\sqrt{x^2 + t^2}} \, dt = \space (*)$ Let's make the change $t = x \cdot \text{ Sh u}$ , $x$ constant, $dt = x \cdot \text{Ch u} \space du$ $(*) = \dfrac d{dx} \int_{Sh^{-1}(x \cdot \text{ Sh}^{2}u)}^{Sh^{-1} x^2} \, \pm \space du =...$ To be continued

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$x >0$, $x = 0$ or $x < 0$ to concrete the ending. This question is giving me questions and suggestions... For example, does $\int_0^{t} \space dt$ make some sense?If the answer is yes, Is it $t$?why?... I have another proof making the change $u = \sqrt{x^2 + t^2} + t$... Even we can think first this question without the derivative $\frac{d}{dx}$, I mean given $x$ possible concrete constants, $x =1$, $x =2$, $x = 100$... and later, seeing the results, infering the last answer... And I'm also considering Pi's advice: Use fundamental theorem of Calculus... To be continued... Can you finish my 2nd proof, and I tell you what I think about your result? Come on, dare...

I think the rest proof 2 should be simple, I'll finish it. I promise.... but you have to say me wheteher I'm right or not... It depends also ifLog in to reply

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@Pi Han Goh @Guillermo Templado help will be welcomed .

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I agree with Pi Hang Goh, you can also make the change $t = x\Sh(u)$ solve , the integral, and later derivate

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I mean $t = x \cdot \text{Sh u}$. I have some problems with my comments..

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Tomorrow, I rewiewed this, and if it's not solved. I'll try to solve it, I think I'm not going to have a lot problemms.. we'll see...

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